Question: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 6} = \dfrac{64}{x - 6}$
Solution: Multiply both sides by $x - 6$ $ \dfrac{x^2}{x - 6} (x - 6) = \dfrac{64}{x - 6} (x - 6)$ $ x^2 = 64$ Subtract $64$ from both sides: $ x^2 - (64) = 64 - (64)$ $ x^2 - 64 = 0$ Factor the expression: $ (x + 8)(x - 8) = 0$ Therefore $x = -8$ or $x = 8$ The original expression is defined at $x = -8$ and $x = 8$, so there are no extraneous solutions.